A convergence problem in distribution theory.

Question

If we identify a locally integrable function with a distribution, then any locally integrable function is a distribution of order 0. Suppose $u_n$ is a constant function for each $n$. If we know that the corresponding distribution $u_n$ in the distribution space converges to a distribution $u$. Then is the distribution $u$ a constant function? To be more precise: $$ u_n\rightarrow u$$ as distributions and $$u_n=C_n$$ where $C_n$ are constants. Is there any constant $C$, such that $$C_n\rightarrow C?$$.In other word is $u$ a constant?

Answer 1

Pick a test function $\varphi$ such that $\varphi \geq 0$ and $\int \varphi \, dx = 1$. Then we have

$$ |\langle C_m, \varphi \rangle - \langle C_n, \varphi \rangle | = \left| (C_m - C_n) \int\varphi \, dx \right| = |C_m - C_n|$$

Since the sequence $( \langle C_n, \varphi \rangle )_{n \geq 1}$ is Cauchy, the same is true for $(C_n)_{n \geq 1}$ and hence $(C_n)$ converges to some constant $C$. Then for any test function $\phi$, we have

$$ \langle C_n, \phi \rangle = C_n \int \phi \, dx \to C \int \phi \, dx = \langle C, \phi \rangle $$

and therefore $C_n \to C$ as distribution.